a.(2b + 1) = 112 = 2⁴.7 = 2⁴.(2.3 + 1), vậy

a = 2⁴ = 16

b = 3

Các bạn có cách khác không?

a) Theo đề, ta có: 

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}\)  và a + b + c =1,5

Theo t/c của dãy tỉ số bằng nhau:

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{2+3+5}=\frac{1,5}{10}=\frac{3}{20}\)

=>a=0,3

    b=0,45

    c=0,75

a) Vì a,b,c tỉ lệ với 2,3,5 

 => \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}\)

Áp dụng t/c dãy tỉ số bằng nhau :

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{2+3+5}=\frac{1,5}{10}=\frac{3}{20}\)

\(\frac{a}{2}=\frac{3}{20}=>a=\frac{3}{20}.2=\frac{3}{10}\)

\(\frac{b}{3}=\frac{3}{20}=>b=\frac{3}{20}.3=\frac{9}{20}\)

\(\frac{c}{5}=\frac{3}{20}=>c=\frac{3}{20}.5=\frac{3}{4}\)

b) 

Áp dụng t/c dãy tỉ số bằng nhau :

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}=\frac{a+2b-3c}{2+6-12}=\frac{-20}{-4}=5\)

\(\frac{a}{2}=5=>a=5.2=10\)

\(\frac{b}{3}=5=>b=5.3=15\)

\(\frac{c}{4}=5=>c=5.4=20\)

c) \(\frac{a}{2}=\frac{b}{3},\frac{b}{5}=\frac{c}{4}\)

 \(\frac{a}{10}=\frac{b}{15},\frac{b}{15}=\frac{c}{12}\)

\(=>\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

Áp dụng t/c dãy tỉ số bằng nhau

\(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{a+b-c}{10+15-12}=\frac{-39}{13}=-3\)

\(\frac{a}{10}=-3=>-3.10=-30\)

\(\frac{b}{15}=-3=>-3.15=-45\)

\(\frac{c}{12}=-3=>-3.12=-36\)

>(a-b)(b-1)+b=15

=>ab-a-b²+b+b=15

=>-b²+2b+ab-a=15

 =>-(b-1)²+a(b-1)=14

=>(b-1){a-(b-1)}=14

=>b-1      1       14   2    7

 a-(b-1)     14      1    7    2

     a          15       13   9   9

     b         2        13     3    8

b)

1: Để a;2a+1;5a-2 lập thành cấp số cộng thì

\(\left[{}\begin{matrix}a=2\left(2a+1+5a-2\right)\\2a+1=2\left(a+5a-2\right)\\5a-2=2\left(a+2a+1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\left(7a-1\right)=a\\2\left(6a-2\right)=2a+1\\5a-2=2\left(3a+1\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}14a-2=a\\12a-4-2a-1=0\\5a-2-6a-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{2}{13}\\a=\dfrac{5}{12}\\a=-4\end{matrix}\right.\)

2:
Để ba số này lập thành cấp số cộng thì 

\(\left[{}\begin{matrix}2b-1=2\left(2b+2-b\right)\\2b=2\left(2b-1+2-b\right)\\2-b=2\left(2b-1+2b\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2b-1=2\left(b+2\right)\left(loại\right)\\2b=2\left(b+1\right)\left(loại\right)\\2-b=2\left(4b-1\right)\end{matrix}\right.\)

=>8b-2=2-b

=>9b=4

=>b=4/9

\(\frac{a^2}{a^2b^2+1}+\frac{b^2}{a^2b^2+1}=\frac{1}{a^2}+\frac{1}{b^2}\)

\(\Leftrightarrow\frac{a^2+b^2}{a^2b^2+1}=\frac{a^2+b^2}{a^2b^2}\)\(\Leftrightarrow a^2b^2\left(a^2+b^2\right)=\left(a^2+b^2\right)\left(a^2b^2+1\right)\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(a^2b^2-a^2b^2-1\right)=0\)

\(\Leftrightarrow a^2+b^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a=0\\b=0\end{cases}}\)

Khúc đầu là: \(\dfrac{1}{a^4+b^2+2b^2}\) hay \(\dfrac{1}{a^4+b^2+2ab^2}\) ??

\(2a^2b\) không phải \(2ab^2\)

Đặt \(\left\{{}\begin{matrix}a-2=x\ge0\\b=y\ge0\end{matrix}\right.\) \(\Rightarrow2y+4=\left(x+2\right)y\Rightarrow xy=4\)

\(P=\dfrac{\sqrt{x^2+2x}}{x+1}+\dfrac{\sqrt{y^2+2y}}{y+1}+\dfrac{1}{x+y+2}\)

\(P=\dfrac{\sqrt{2x\left(x+2\right)}}{\sqrt{2}\left(x+1\right)}+\dfrac{\sqrt{2y\left(y+2\right)}}{\sqrt{2}\left(y+1\right)}+\dfrac{1}{x+1+y+1}\)

\(P\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{3x+2}{x+1}+\dfrac{3y+2}{y+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}\right)\)

\(P\le\dfrac{1}{2\sqrt{2}}\left(3-\dfrac{1}{x+1}+3-\dfrac{1}{y+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}\right)\)

\(P\le\dfrac{3\sqrt{2}}{2}-\dfrac{\sqrt{2}-1}{4}\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}\right)\)

Ta có:

\(\dfrac{1}{x+1}+\dfrac{1}{y+1}=\dfrac{x+y+2}{xy+x+y+1}=\dfrac{x+y+2}{x+y+5}=1-\dfrac{3}{x+y+5}\ge1-\dfrac{3}{2\sqrt{xy}+5}=\dfrac{2}{3}\)

\(\Rightarrow P\le\dfrac{3\sqrt{3}}{2}-\dfrac{\sqrt{2}-1}{4}.\dfrac{2}{3}=...\)

Dấu "=" xảy ra khi \(x=y=2\) hay \(\left(a;b\right)=\left(4;2\right)\)

Ta có: 

+\(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\)

\(\Rightarrow\dfrac{1}{a}\ge\dfrac{2b-1}{2b+1}+\dfrac{3c-1}{3c+2}\ge2\sqrt{\dfrac{\left(2b-1\right)\left(3c-1\right)}{\left(2b+1\right)\left(3c+2\right)}}\left(1\right)\)

+\(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\)

\(\Rightarrow\dfrac{2}{2b+1}\ge\dfrac{a-1}{a}+\dfrac{3c-1}{3c+2}\ge2\sqrt{\dfrac{\left(a-1\right)\left(3c-1\right)}{a\left(3c+2\right)}}\left(2\right)\)

+\(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\)

\(\Rightarrow\dfrac{3}{3c+2}\ge\dfrac{a-1}{a}+\dfrac{2b-1}{2b+1}\ge2\sqrt{\dfrac{\left(a-1\right)\left(2b-1\right)}{a\left(2b+1\right)}}\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow6\ge8\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\)

\(\Rightarrow P=\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\le\dfrac{3}{4}\)

\(\Rightarrow P_{max}=\dfrac{3}{4}\) đạt tại \(a=\dfrac{3}{2};b=1;c=\dfrac{5}{6}\)